Which of the following solutions has the lowest pH? 0.002-M HClO4, 0.20-M CH3COOH, 0.02-M HCl, 2.0-M NaCl, 0.200-M HCOOH (I asked this before but I am unsure if the answer I received is correct so I am resubmitting it.)
a) 0.002 M HClO4 is a strong acid and ionizes completely.
HClO4 (aq) --------> H+ (aq) + ClO4- (aq)
[H+] = [HClO4] = 0.002 M.
pH = -log (0.002)
= 2.699 ≈ 2.7
b) CH3COOH is a weak acid and ionizes partially.
CH3COOH (aq) <=====> H+ (aq) + CH3COO- (aq) Ka = 1.8*10-5
Ka = [H+][CH3COO-]/[CH3COOH]
=====> 1.8*10-5 = (x)(x)/(0.20 – x)
where x M is the equilibrium concentration of H+.
Since Ka is small, assume x << 0.20 M and write,
1.8*10-5 = x2/0.20
=====> x2 = 1.8*10-5*0.20
=====> x2 = 3.6*10-6
=====> x = 1.8974*10-3
[H+] = 1.8974*10-3 M
pH = -log [H+]
= -log (1.8974*10-3 M)
= 2.7218 ≈ 2.7
c) 0.02 M HCl is a strong acid and ionizes completely.
HCl (aq) --------> H+ (aq) + Cl- (aq)
[H+] = [HCl] = 0.02 M.
pH = -log (0.02)
= 1.699 ≈ 1.7
d) NaCl is a neutral salt and hence, the pH = 7.0.
e) HCOOH is a weak acid and ionizes partially.
HCOOH (aq) <=====> H+ (aq) + HCOO- (aq) Ka = 1.8*10-4
Ka = [H+][HCOO-]/[HCOOH]
=====> 1.8*10-4 = (x)(x)/(0.20 – x)
where x M is the equilibrium concentration of H+.
Since Ka is small, assume x << 0.20 M and write,
1.8*10-5 = x2/0.20
=====> x2 = 1.8*10-4*0.20
=====> x2 = 3.6*10-5
=====> x = 6.0*10-3
[H+] = 6.0*10-3 M
pH = -log [H+]
= -log (6.0*10-3 M)
= 2.2218 ≈ 2.2
Therefore, 0.02 M HCl will have the lowest pH (ans).
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